TITLE: Does $(\cos t , \sin (\sqrt{2}t))$ traces all the points of a square.
QUESTION [15 upvotes]: The following parametric equation recently across my mind recently,
$$x = \cos (t)$$
$$y = \sin (\sqrt{2} t)$$
And graphing on my calculator shows it eventually traces out a "square". What I am wondering is if this function actually traces out all the points within this square for $t \in \mathbb{R}$, or some region of it as the corners will never be reached.

REPLY [3 votes]: For a fixed $x \in [-1,1]$, there are countably many $t$ that satisfy the parametric equations. Let $T$ denote such values of $t$. $y$ has period $\sqrt2\pi$.

Convince yourself that $T \pmod{\sqrt2\pi}$ is dense in $[0,\sqrt2\pi]$. (1, 2, 3)
Then, convince yourself that $\sin \sqrt 2T$ is dense in $[-1,1]$. (limit commutes with continuous function)
However, convince yourself that $\sin \sqrt 2T$ only has countably many numbers, while any particular vertical strip of the square has uncountably many numbers, so not all points are reached. (see comment)
Conclude that since $x$ is arbitrary, the set of points $(x,y)$ in the graph is dense in $[-1,1]^2$, but does not contain every point in $[-1,1]^2$.